The distance of the point A(3,-4,5) from the plane 2 x+5 y-6 z=16 measured along the line x 2 = y 1 = z -2 is

Step-by-step solution

The distance from A(3,-4,5) to the plane 2x+5y-6z=16 measured along the direction (2,1,-2) is determined by finding the intersection point of the line through A parallel to this direction with the plane. The parametric equations of the line are x=3+2 , y=-4+ , z=5-2 . Substituting into the plane equation yields: 2(3+2 )+5(-4+ )-6(5-2 )=16 21 -44=16 = 60 21 = 20 7 The intersection point coordinates are: x_B=3+2 20 7 = 61 7 y_B=-4+ 20 7 = -8 7 z_B=5-2 20 7 = -5 7 The distance AB is calculated as: ( 61 7 -3 )^2+ ( -8